Coordinate Geometry Test

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Coordinate Geometry Test - 18

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Question 1
1 / -0

The area of a triangle whose vertices are $$(a, c+a), (a, c) $$ and $$(-a, c-a) $$ are

A
$$\displaystyle a^{2}$$

B
$$\displaystyle b^{2}$$

C
$$\displaystyle c^{2}$$

D
$$\displaystyle a^{2}+c^{2}$$

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
Hence, area of the triangle with given vertices is $$ \left| \frac { { a }(c-c+a)+a(c-a-c-a)-a(c+a-c) }{ 2 } \right| = \left| \frac { { a }^{ 2 }-2{ a }^{ 2 }-{ a }^{ 2 } }{ 2 } \right| = { a }^{ 2 } $$
Question 2
1 / -0

If the area of the triangle formed by the points $$(-2,3), (4,-5)$$ and $$(-3,y)$$ is 10 square units then $$y =$$

A
$$\dfrac{19}{3}$$

B
$$\dfrac{21}{3}$$

C
$$\dfrac{23}{3}$$

D
$$\dfrac{25}{3}$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
$$\displaystyle \text{Area} =\frac{1}{2}\left [ -2\left ( -5-y \right )+4\left ( y-3 \right )-3\left ( 3+5 \right ) \right ]$$
$$\displaystyle =\frac{1}{2}\left ( 10+2y+4y-12-24 \right )$$
$$\displaystyle =\frac{1}{2}\left ( 6y-26 \right )=3y-13$$
$$\displaystyle 3y-13=10\ \mathrm{sq. unit}$$
$$\displaystyle y=\frac{23}{3}$$
Question 3
1 / -0

The midpoint of the line segment with endpoints $$(-6,4)$$ and $$(8,2)$$ is

A
$$(3,1)$$

B
$$(1,3)$$

C
$$(2,6)$$

D
$$(3,3)$$

Solution

Midpoint = $$\displaystyle \left ( \frac{-6+8}2,{\frac{4+2}{2}} \right )=\left ( 1, 3 \right )$$
Question 4
1 / -0

If the area of the triangle formed by $$ (-2,5), (x,-3) $$ and $$(3,2)$$ is $$14 $$ square units, then $$x=$$ ____.

A
1

B
2

C
-2

D
-1

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
$$\therefore \displaystyle Area=\frac{1}{2}\left [ -2\left ( -3-2 \right )+x\left ( 2-5 \right )+3\left ( 5+3 \right ) \right ]$$
$$\displaystyle =\frac{1}{2}\left [ 10-3x+24 \right ]=\frac{34-3x}{2}$$
$$\displaystyle \therefore \frac{34-3x}{2}=14$$ or x = 2
Question 5
1 / -0

A(4,0) and B(5,0) are two points If AB is produced to C such that AB=BC then the coordinates of C are

A
(6,0)

B
(9,0)

C
(11,0)

D
(12,0)

Solution

From the figure C = (6, 0)
Question 6
1 / -0

In the diagram M(2,3) is the midpoint of AB The coordinates of A and B are respectively

A
(4,0) and (0,6)

B
(3,0) and (0,2)

C
(4,0) and (0,4)

D
(6,0) and (0,6)

Solution

Let A be (x, 0) and B be (0, y) Then
$$\displaystyle M=\left ( \frac{0+x}{2},\frac{y+0}{2} \right )=\left ( 2,3 \right )$$
or x = 4, y = 6
Question 7
1 / -0

In the diagram, $$M$$ is the midpoint of $$PQ$$. Find the value of $$k$$.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$5$$

Solution

Applying mid point formula for point $$P(-3,1)$$ and $$Q(5,k)$$ and equating the coordinates to point $$M$$ we get,

$$\dfrac{5+(-3)}{2}=1$$

$$\dfrac{k+1}{2}=2$$

$$\therefore\Rightarrow k=3$$
Question 8
1 / -0

The midpoint of the line segment between P$$\displaystyle _{1}$$ (x, y) and P$$\displaystyle _{2}$$ (-2, 4) is P$$\displaystyle _{m}$$ (2, -1). Find the unknown coordinate.

A
(6, -5)

B
(5, -6)

C
(6, -6)

D
(-6, 6)

Solution

Given,
$$P_m(2,-1), P_1(x,y)$$ and $$P_2(-2,4)$$

$$(2,-1)=\left(\dfrac{x+(-2)}{2}, \dfrac{y+4}{2}\right)$$ ..... By midpoint formula

$$\therefore 2=\dfrac{x+(-2)}{2}$$
$$=>4=x+(-2)$$
$$=>x=6$$

And,
$$-1=\dfrac{y+4}{2}$$
$$=>-2=y+4$$
$$=>y=-6$$

$$\therefore (x,y)=(6,-6)$$
Question 9
1 / -0

In the following figure, what is the midpoint of $$\overline {UV}$$?

A
$$\left (-1, \dfrac {1}{2}\right )$$

B
$$\left (-1, \dfrac {3}{2}\right )$$

C
$$\left (-1, -\dfrac {1}{2}\right )$$

D
$$\left (-2, \dfrac {1}{2}\right )$$

E
$$\left (-2, -\dfrac {3}{2}\right )$$

Solution

Given are the two points of the line $$VU$$
Let point on $$V$$ be$$\left( { x }_{ 1 },y_{ 1 } \right) $$ $$=(1,8)$$
Let point on $$U$$ be $$\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ $$=(-3,-7)$$
We know the midpoint formula is
$$\left( { x }_{ m },{ y }_{ m } \right) =\left[ \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right] $$
Putting the values of in the formula, we get
$$\left( { x }_{ m },{ y }_{ m } \right) =\left[ \dfrac { 1+\left( -3 \right) }{ 2 } ,\dfrac { 8+\left( -7 \right) }{ 2 } \right] $$
$$\left( { x }_{ m },{ y }_{ m } \right) =\left[ \dfrac { -2 }{ 2 } ,\dfrac { 1 }{ 2 } \right] $$
$$\left( { x }_{ m },{ y }_{ m } \right) =\left[ -1,\dfrac { 1 }{ 2 } \right] $$
Question 10
1 / -0

In the xy-plane, find the mid point of the line segment joining the points $$\left( 5,9 \right) $$ and $$\left( 7,11 \right) $$.

A
$$(1.5, 2)$$

B
$$(6, 10)$$

C
$$(5.5, 5)$$

D
$$(6, -3.5)$$

Solution

Given line segment point $$(5,9)$$ and $$(7,11)$$, then mid point as per section formula:
$$(x,y)=$$ $$\left ( \dfrac{7+5}{2} ,\dfrac{11+9}{2} \right )$$
$$=$$ $$\left ( \dfrac{12}{2} ,\dfrac{20}{2}\right)$$
$$=$$ $$ (6,10)$$

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Coordinate Geometry Test - 18

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Coordinate Geometry Test - 18

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Question 1

$$\displaystyle a^{2}$$

$$\displaystyle b^{2}$$

$$\displaystyle c^{2}$$

$$\displaystyle a^{2}+c^{2}$$

Solution

Question 2

$$\dfrac{19}{3}$$

$$\dfrac{21}{3}$$

$$\dfrac{23}{3}$$

$$\dfrac{25}{3}$$

Solution

Question 3

$$(3,1)$$

$$(1,3)$$

$$(2,6)$$

$$(3,3)$$

Solution

Question 4

1

2

-2

-1

Solution

Question 5

(6,0)

(9,0)

(11,0)

(12,0)

Solution

Question 6

(4,0) and (0,6)

(3,0) and (0,2)

(4,0) and (0,4)

(6,0) and (0,6)

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$5$$

Solution

Question 8

(6, -5)

(5, -6)

(6, -6)

(-6, 6)

Solution

Question 9

$$\left (-1, \dfrac {1}{2}\right )$$

$$\left (-1, \dfrac {3}{2}\right )$$

$$\left (-1, -\dfrac {1}{2}\right )$$

$$\left (-2, \dfrac {1}{2}\right )$$

$$\left (-2, -\dfrac {3}{2}\right )$$

Solution

Question 10

$$(1.5, 2)$$

$$(6, 10)$$

$$(5.5, 5)$$

$$(6, -3.5)$$

Solution

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